Q. 4.83 - Intermediate Algebra

Question

Solve the system of equations using a matrix: $\left\{\begin{array}{l}2x-5y+3z=8\\ 3x-y+4z=7\\ x+3y+2z=-3\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(6,-1,-3\right)$.

Step 1. Given information.

Consider the given system of equations,

$\left\{\begin{array}{l}2x-5y+3z=8\\ 3x-y+4z=7\\ x+3y+2z=-3\end{array}\right.$

Step 2. Write in augmented form.

The augmented matrix for the given system of equations is

$\left[\begin{array}{cccr}2& -5& 3& 8\\ 3& -1& 4& 7\\ 1& 3& 2& -3\end{array}\right]$

Step 3. Apply row operations.

Apply  $\frac{R_1}{2}\to R_1$ and $R_2-3\times R_1\to R_2$,

$\left[\begin{array}{cccr}1& -\frac{5}{2}& \frac{3}{2}& 4\\ 0& \frac{13}{2}& -\frac{1}{2}& -5\\ 1& 3& 2& -3\end{array}\right]$

Apply $R_3-R_1\to R_3$ and $R_2\times \frac{2}{13}\to R_2$,

$\left[\begin{array}{cccr}1& -\frac{5}{2}& \frac{3}{2}& 4\\ 0& 1& -\frac{1}{13}& -\frac{10}{13}\\ 0& \frac{11}{2}& \frac{1}{2}& -7\end{array}\right]$

Apply $R_3\times \frac{13}{12}\to R_3$,

$\left[\begin{array}{cccr}1& -\frac{5}{2}& \frac{3}{2}& 4\\ 0& 1& -\frac{1}{13}& -\frac{10}{13}\\ 0& 0& 1& -3\end{array}\right]$

Now, the matrix is in row-echelon form.

Step 4. Write in system of equations.

Writing the corresponding system of equations,

$$\begin{gathered} x-\frac{5}{2}y+\frac{3}{2}z=4 ...... \left(i\right) \\ y-\frac{1}{13}z=-\frac{10}{13} ...... \left(ii\right) \\ z=-3 ...... \left(iii\right) \end{gathered}$$

Substitute the value of $z$ in equation (ii),

$$\begin{gathered} y-\frac{1}{13}\times \left(-3\right)=-\frac{10}{13} \\ y=-\frac{10}{13}-\frac{3}{13} \\ y=-\frac{13}{13} \\ y=-1 \end{gathered}$$

Substitute the value of $y,z$ in equation (i),

$$\begin{gathered} x-\frac{5}{2}\times \left(-1\right)+\frac{3}{2}\times \left(-3\right)=4 \\ x+\frac{5}{2}-\frac{9}{2}=4 \\ x-\frac{4}{2}=4 \\ x=6 \end{gathered}$$

Step 5. Check the answers.

Substitute the values $x,y,z$ in equation (i),

$$\begin{gathered} 6-\frac{5}{2}\left(-1\right)+\frac{3}{2}\left(-3\right)=4 \\ 6+\frac{5}{2}-\frac{9}{2}=4 \\ \frac{8}{2}=4 \\ 4=4 \end{gathered}$$

This is true.

Substitute the values $y,z$ in equation (ii),

$$\begin{gathered} -1-\frac{1}{13}\left(-3\right)=-\frac{10}{13} \\ -1+\frac{3}{13}=-\frac{10}{13} \\ -\frac{10}{13}=-\frac{10}{13} \end{gathered}$$

This is also true.